Chapter 3.6, Problem 42E

  $\left(a\right)$

To determine

To show: The function $f\left(x\right)=2x+\cos x$ is one-to-one.

Answer to Problem 42E

  $f\left(x\right)=2x+\cos x$ is one-to-one.

Explanation of Solution

Given information:

The function $f\left(x\right)=2x+\cos x$ .

Formula used:

A function f is said to be one-one function if it is an increasing function and defined on real numbers.

Consider the function $f\left(x\right)=2x+\cos x$ .

Differentiate both sides with respect to x ,

  $\begin{array}{c}\frac{d}{dx}\left[f\left(x\right)\right]=\frac{d}{dx}\left[2x+\cos x\right]\\ f\text{'}\left(x\right)=\frac{d}{dx}\left[2x\right]+\frac{d}{dx}\left[\cos x\right]\\ f\text{'}\left(x\right)=2-\sin x\end{array}$

Recall that range of sine trigonometric function is $\left[-1,1\right]$ . So,

  $\begin{array}{c}-1\le \sin x\le 1\\ -1\le -\sin x\le 1\\ 1\le 2-\sin x\le 3\end{array}$

The derivative of the function is positive, so the function is an increasing function.

Recall that a function f is said to be one-one function if it is an increasing function and defined on real numbers.

Since, the function $f\left(x\right)=2x+\cos x$ is increasing and defined on real values so it is an increasing function.

Hence, it is proved that thefunction $f\left(x\right)=2x+\cos x$ is one-to-one.

  $\left(b\right)$

To determine

To calculate: The value of the expression $f^{-1}\left(1\right)$ .

Answer to Problem 42E

The value of the expression $f^{-1}\left(1\right)$ is $0$ .

Explanation of Solution

Given information:

The function $f\left(x\right)=2x+\cos x$ .

Formula used:

If $f$ is a one-to-one and onto differentiable function and the inverse of the function $f^{-1}$ exist.

Calculation:

Consider the provided function $f\left(x\right)=2x+\cos x$ . To evaluate the value of $f^{-1}\left(1\right)$ . Denote $f^{-1}\left(1\right)=x$ this implies that $f\left(x\right)=1$ . Construct the graphs of the function $f\left(x\right)=2x+\cos x$ and $f\left(x\right)=1$ in the same graphing window. Observe the point where both the graphs interest each other.

The red curve depicts the graph of $f\left(x\right)=2x+\cos x$ and blue line depicts the function $f\left(x\right)=1$ . The point where they intersect is $\left(0,1\right)$ . That is $f\left(0\right)=1$ . Therefore, $f^{-1}\left(1\right)=0$ .

Thus, the value of the expression $f^{-1}\left(1\right)$ is $0$ .

  $\left(c\right)$

To determine

To calculate: The value of the expression $\left(f^{-1}\right)\text{'}\left(1\right)$ .

Answer to Problem 42E

The value of the expression $\left(f^{-1}\right)\text{'}\left(1\right)$ is $\frac{1}{2}$ .

Explanation of Solution

Given information:

The function $f\left(x\right)=2x+\cos x$ .

Formula used:

If $f$ which is a one-to-one differentiable function and the inverse of the function $f^{-1}$ is also differentiable then $\left(f^{-1}\right)\text{'}\left(x\right)=\frac{1}{f\text{'}\left(f^{-1}\left(x\right)\right)}$ provided that denominator is not equal to 0.

Calculation:

Consider the provided function $f\left(x\right)=2x+\cos x$ .

Also derivative of the function is $f\text{'}\left(x\right)=2-\sin x$ .

Now, derivative of the function at $x=0$ is,

  $\begin{array}{c}f\text{'}\left(0\right)=2-\sin \left(0\right)\\ =2-0\\ =2\end{array}$

It is found that $f^{-1}\left(1\right)=0$ .

Recall that if $f$ which is a one-to-one differentiable function and the inverse of the function $f^{-1}$ is also differentiable then $\left(f^{-1}\right)\text{'}\left(x\right)=\frac{1}{f\text{'}\left(f^{-1}\left(x\right)\right)}$ provided that denominator is not equal to 0.

So, to evaluate the value of $\left(f^{-1}\right)\text{'}\left(1\right)$ apply the above formula,

  $\left(f^{-1}\right)\text{'}\left(1\right)=\frac{1}{f\text{'}\left(f^{-1}\left(1\right)\right)}$

Since, $f^{-1}\left(1\right)=0$ and $f\text{'}\left(0\right)=2$ , so

  $\begin{array}{c}\left(f^{-1}\right)\text{'}\left(1\right)=\frac{1}{f\text{'}\left(f^{-1}\left(1\right)\right)}\\ =\frac{1}{f\text{'}\left(0\right)}\\ =\frac{1}{2}\end{array}$

Thus, the value of the expression $\left(f^{-1}\right)\text{'}\left(1\right)$ is $\frac{1}{2}$ .